Question

$300\text{J}$ of work done in sliding a $2\text{kg}$ block up the inclined plane of height $10\text{m}$ with constant velocity. The work done against friction is (Take $g=10{\text{m/s}}^2$)

• A. $100\text{J}$
• B. Zero $\text{J}$
• C. $200\text{J}$
• D. $500\text{J}$

Answer 1

The correct option is A $100\text{J}$

Let the applied force $F$ be parallel to the inclined plane. When it pulls the block up, kinetic friction acts down to oppose the relative motion.

The work done by the force to displace the block through a distance $l$ along the inclined plane is given as
$W=F.s=Fl$


$l=hcosec\theta$
$\Rightarrow W=Fhcosec\theta \dots \dots \left(2\right)$
Since the block slides up with constant velocity, therefore the net force acting on the block
$\sum F_x=0;\sum F_y=0\Rightarrow mg\sin \theta +f_k=F\dots \dots \left(3\right)$

Using 2 and 3 we get
$W=(mg\sin \theta +f_k)hcosec\theta$

$=mgh+f_{k}h\text{cosec}\theta$

$=300\text{J}$ ......(given)

$\therefore W_k(\text{Work done against friction)}=f_{k}h\text{cosec}\theta =300\text{J}-mgh=300\text{J}-(2\times 10\times 2)\text{J}=300\text{J}-200\text{J}=100\text{J}$